John L.
asked • 07/20/14Probability to get a blue ball
Marble produced at a certain factory has a 20% chance of being blue. A box contains 10 marbles.
(a)Find the probability that exactly 2 balls out of the 10 balls in the box are blue.
(b) Find the probability that at least 8 balls out of the 10 balls in the box are blue.
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Dattaprabhakar G. answered • 07/20/14
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John L.
No its not like that. I tried that question but I am not sure if that it the right way to do it. Please tell me if it is or it is not
(a) 10C2*(0.2)2 * (0.8)2
(b)10C8*(0.2)8*(0.8)2 + 10C9*(0.2)9*(0.8)1 + 10C10*(0.2)10
I am not sure if i need to select those blue balls before by the formula 10Cn
Please help me with it.
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07/20/14
John L.
Hi,
It is not like that. I have got a theory but I am not sure as the final answer seem to low. This is what I understand.
(a) (0.2)2 * (0.8)8
(b) (0.2)8 * (0.8)2 + (0.2)9 * (0.8)1 + (0.2)10 * (0.8)0
Is this answer right ? Do i need to use the nCr function too somewhere ?
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07/20/14
Dattaprabhakar G.
John:
Thanks for the response. I shall give the (b) part separately.
(a) Yes, you must use the "super_n_C_sub_r" expression. The reason for that is that the event "exactly 2 balls out of the 10 balls in the box are blue." contains many possibilities of the ORDER in which the two blue balls occur in the set of 10 marbles. Denoting by B a blue marble and by O any other marble, the event "exactly 2 balls out of the 10 balls in the box are blue." consists of elementary events such as BBOOOOOOOO, BOBOOOOOOO, etc. The number of such elementary events is "10 Choose 2" = "super_10_C_sub_2"= (10!)/(2!)(8!) = (10 x 9 x 8!)/2 x (8!) = 45. You have calculated the probability of the elementary event that there are exactly 2 blue balls IN A SPECIFIC ORDER, assuming that the marbles are produced randomly and independently.
So, the probability that exactly 2 balls out of the 10 balls in the box are blue equals 45 x (0.2)2 x (0.8)8.
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07/20/14
Dattaprabhakar G.
Hi John: (Excuse me for using marbles instead of balls!)
To continue with my earlier comment, yes, in the (b) part too, you have to multiply by the number of arrangements. Thus "exactly 9 blue marbles in a set of 10" (which is a part of "at least 8 marbles out of the 10 marbles in the box are blue."), as you have noted, consists of "10 Choose 9" = (10!) / (9!)(1!) = 10 arrangements each having the probability (0.2)9(0.8) assuming that the marbles are produced randomly and independently with the same chance throughout of either being BLUE or OTHER THAN BLUE.
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07/20/14
John L.
Thank you Dattaprabhakar for the elaborated explanation. Now I definitely have a much better understanding.
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07/20/14
Dattaprabhakar G.
John:
You seem to be a talented student. So here is something for you to think about.
Go back to your original problem of a factory producing marbles. But, instead of producing, the factory just PACKS 10 marbles in a box, choosing them randomly but without replacement (just like dealing 20 playing cards from a well-shuffled deck of 200 cards) from an inventory of 200 marbles out of which 40 (20%) are BLUE and the remaining 160 are of a different color.
Answer the same questions of yours, Questions (a) and (b), under the above scenario. If you get stuck, post a comment. I will be happy to help. (BTW, leave your answers in terms of the nCr coefficients. No need to simplify. I trust your computational ability. )
Dattaprabhakar
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07/20/14
John L.
This is what I think the answer should look like. I am not sure about part b as I don't know in this particular case how to take 'without replacement' into account.
(a) 40C2 * 160C8 * (0.2)^2 * (0.8)^8
(B)40C8 * 160C2 * (0.2)^8 * (0.8)^2 + 40C9 * 160C1 * (0.2)^9 * (0.8)^1 + 40C10 * (0.2)^10
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07/21/14
Dattaprabhakar G.
John:
No, your answers are not correct, but you get an A for the prompt effort. Do not be disheartened, because I was not really expecting you to know the stuff. I will explain in detail how the correct answer is found, but please give me until Monday (07/21)
night.
Dattaprabhakar ("Dr.G.")
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07/21/14
John L.
Sure
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07/21/14
Dattaprabhakar G.
John: This comment is sent to you in parts. Please look at all the four parts and some material that follows at the end.
1) First note that when you are selecting a batch of 10 marbles, even though randomly, sampling with replacement does not make any sense; you can not have the same marble “twice”, virtually, in the batch, which CAN happen if the marble
is replaced in the inventory. In many real life situations also, sampling with replacement is a likely waste of money because you may get the same unit of observation to look at. It also involves duplicity of information. Sampling with replacement is assumed
for mathematical convenience more than anything else. It gives rise to INDEPENDENT, IDENTICALLY DISTRIBUTED observations. ..........
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07/21/14
Dattaprabhakar G.
2) Second, note that your expression (0.2)2(0.8)8 , for example, assumes independence (so that you can multiply probabilities) and that the probability of selecting a blue or a non-blue marble remains constant (0.2 for blue and
0.8 for non-blue) when each marble is selected. These assumptions are violated when you sample without replacement.
For example, suppose that a non-blue marble is selected on the first draw (chance 160/200 = 0.8). The probability of selecting a non-blue marble on the second draw depends on what happened on the first draw. The draws are NOT independent. Given that the first marble drawn is non-blue, the probability of drawing a non-blue marble on the second draw is 159/199 = 0.799. Hence the probability of "success" does not remain constant from draw to draw either.
For this reason, you can not write expressions of the form (0.8)x (0.2)y as probabilities of elementary events.
For example, suppose that a non-blue marble is selected on the first draw (chance 160/200 = 0.8). The probability of selecting a non-blue marble on the second draw depends on what happened on the first draw. The draws are NOT independent. Given that the first marble drawn is non-blue, the probability of drawing a non-blue marble on the second draw is 159/199 = 0.799. Hence the probability of "success" does not remain constant from draw to draw either.
For this reason, you can not write expressions of the form (0.8)x (0.2)y as probabilities of elementary events.
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07/21/14
Dattaprabhakar G.
John:
Did you get everything, all 4 parts + some helpful web-link material? Please confirm.
Dr.G.
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07/21/14
Dattaprabhakar G.
John, I do not see my part 3 in the comments. Have you received it? I have repeatedly sent it. Dattaprabhakar
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07/22/14
John L.
Dr G,
Sorry for the delay in response. Thank you for the explanation but I can only see 2 comments here and you said you had sent 4.
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07/25/14
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Dattaprabhakar G.
By the way, what I have givrn you is sn introduction to the “Hypergeometric Distibution”.
A useful web-link is http://stattrek.com/online-calculator/hypergeometric.aspx where you can calculate the probabilities. Go to the web-link and enter Population size = 200, Number of successes in the population = 40 (blue marbles), Sample size = 10, Number os successes in the sample = 2 (or 8, or 9, or 10). Click on “Calculate” at the lower right corner. You will get the desired probability and a whole bunch of other quantities.
07/21/14